groups of order 1575 with a unique Sylow 3-subgroup

Question

I am trying to prove (using a different method) that if $|G| = 1575$ and $P_3 \in Syl_3(G)$ is unique, then $P_5$ is also normal in $G$.

I already have 2 proofs of this. The first one uses the automorphism group of $P_3$ as a consequence and the other one uses products of the Sylow subgroups. (See If $G$ is a group of order $1575$ with normal Sylow $3$ subgroup, then show that Sylow $5$ and $7$ subgroups are both normal.).

Please do not give these methods as hints.

I have found the following facts about G.

1. $n_5 = 1$ or $21$
2. $n_7 = 1, 15$

• if $n_7 = 225$, then the contradiction is by showing that this forces all elements to be of order 25, 9, 7, 5, 3, or 1, but then we show that there exists an order 15 subgroup (which is always cyclic). Thus, we have an order 15 element which is the contradiction.

3. If $n_5 = 21$, then we are forced to have $|N_G(P_5)| = 3 \times5^2$. Also, if H is the cyclic group of order 3 that is contained in $N_G(P_5)$ then $HP_5 = N_G(P_5)$.

A. Note that $P_3$ is order 9. Thus, it is isomorphic to either $Z_9$ or $Z_3\times Z_3$.

• If $P_3 \cong Z_9$, then H is the unique subgorup of order 3 (by classification of cyclic groups). Thus, $H$ char $P_3$ and $P_3 \unlhd G$ gives $H \unlhd G$.

Consider now the group $\bar{G} = G/H$ which has order $3\times 5^2\times 7$. let $K = N_G(P_5)$. We have that $\bar{K}\in Syl_5(\bar{G})$.

B. $\bar{K} \unlhd \bar{G}$

• because $n_5({\bar{G}})$ is forced to be either 1 or 21, we only need to show it can't be 21. Suppose it is. This forces $N_{\bar{G}}(\bar{K})$ = $5^2$.
• Since $|\bar{K}| = 5^2$, $\bar{K}$ is abelian. Thus, $\bar{K} \leq C_{\bar{G}}(\bar{K}) \leq N_{\bar{G}}(\bar{K})$ and $\bar{K} = N_{\bar{G}}(\bar{K})$ forces $\bar{K} =C_{\bar{G}}(\bar{K}) = N_{\bar{G}}(\bar{K})$
• Thus, $Z(\bar{G}) \leq C_{\bar{G}}(\bar{K}) = \bar{K}$. The other side is $P_3 \leq Z(\bar{G})$. This is a contradiction because of divisibility issues. (RHS is 5^2 and the left side is 3 which is a contradiction by Lagrange Theorem.)

C. By Lattice Isomorphism Theorem, $K\unlhd G$. This means, $\forall g \in G$, $N_G(gP_5g^{-1}) = gN_G(P_5)g^{-1} = N_G(P_5)$. This means that all Sylow 5-subgroups of G have the same normalizer in G.

D. Let $A,B \in Syl_5(G)$. Then $A \in N_G(A), B \in N_G(B),$ and $N_G(A) = N_G(B)$ forces $A,B \in N_G(P_5)$. Also $A \in N_G(A) = N_G(B)$ makes $AB$ into a group. Thus, $AB \leq N_G(P_5)$.

E. But $|N_G(P_5)| = 3 \times 5^2$ and $|AB| = \frac{|A||B|}{|A \cap B|} = \frac{5^4}{|A \cap B|}$. By Lagrange Theorem, we are forced to have $|AB| = 1, 5, or 5^2$. The only possibility is $|AB| = 5^2$ which forces $|A \cap B| = 5^2$. This means $ A = B$.

F. Thus, there is a unique Sylow 5-subgroup, which contradicts our assumption $n_5 = 21$.

Now, I am stuck. How do I prove the case when $P_3 \cong Z_3 \times Z_3$ supposing we do not use the idea of automorphism groups.

Answer 1

This is actually automorphism group in disguise, but certainly we can do without mentioning it.

Note that the normal $P_3\cong C_3\times C_3$ has 4 subgroups of order 3, and so we have homomorphism $H\to S_4$ for any subgroup $H$ of $G$, permuting these 4 subgroups by conjugation. In particular, since there are no element of order 5 (or 7) in $S_4$, $P_5$ (or $P_7$) must act trivially, i.e., they normalises each subgroup of order 3. But that is enough to conclude $n_5,n_7$ cannot be divisible by 3 (because the only way to fix a group of order 3 is either trivial or swapping the two order 3s but 5,7 are odd).